# Simplicial complex made of central idempotents of an algebra

Let $A$ be an algebra, say over $\mathbb{C}$ and finite-dimensional, but not necessary semisimple. I have the strong feeling, which I would like to prove and use, about the following rather natural approach.

Consider the set $S$ of central idempotents in $A$, then I want to turn the set of central idempotents $\Delta$ into an abstract simplicial complex. Two idempotents $e_1,e_2$ are connected iff $e_1\in e_2A$, simplices consist of completely connected subset and and the highest-dimensional simplices hence are flags associated to a decomposition of $A$ into central primitive idempotents. This is much how one defines buildings as simplicial complexes e.g. of subspaces of a finite vector space....

Is this OK? I'm not familiar with calculating with central prinimtive idempotents - especially I'm concerned, if any decomposistion into central primitive idempotents has the same cardinality??

..if yes,it seems so natural, something similar ought to exist (surely more advanced ;-) - any reference? I would probably also suffice with a related construction and/or I want to find out more established facts about such an "algebraic simplicial complex / building"...Thanks for your help in advance.

The central idempototents of a finite dimensional algebra form a finite Boolean algebra with 1 as the max and the central primitive idempotents as the atoms. The decomposition of 1 into central idempotents is thus unique. So the central idempotents are the face lattice of a simplex. The order is $e\leq f$ if $e\in fA$. Your simplicial complex would be the order complex of this Boolean algebra and so would be the barycentric subdivision of a simplex.
Added details. The boolean algebra operations are given by $e\wedge f=ef$, $e\vee f=e+f-ef$ and $\neg e=1-e$. The finiteness follows, for example, because one can look at the regular representation of the algebra $A$ by matrices and observe that the central idempotents form a commutative semigroup of idempotent matrices. Such a semigroup is simultaneously diagonalizable and there are only $2^n$ diagonal idempotent $n\times n$ matrices.